Question

Hydroelectric generators at Hoover Dam produce a maximum current of 8.00×10³ A at 250kV. What is the power output (in W)? (b) The water that powers the generators enters and leaves the system at low speed (thus its kinetic energy does not change) but loses 160 m in altitude. How man cubic meters per second are needed, assuming 87.0% efficiency? (Assume the density of water is 1,000 kg/m ?. ) m³/s

Answer 1

The power output of the hydroelectric generators at Hoover Dam is 2 GW. Approximately 11,920 cubic meters per second of water are needed, assuming 85.0% efficiency.

Step-by-step explanation:

To calculate the power output of the hydroelectric generators at Hoover Dam, we can use the formula:

Power (P) = Current (I) x Voltage (V)

Given that the maximum current is 8.00 × 10³ A and the voltage is 250 kV, we need to convert the voltage to volts:

250 kV = 250,000 volts

Plugging in the values, we get:

Power = 8.00 × 10³ A x 250,000 volts = 2 x 10^9 watts = 2 GW

Therefore, the power output of the hydroelectric generators at Hoover Dam is 2 GW.

To calculate the flow rate of water needed, we can use the formula:

Power = Efficiency x Flow Rate x Density x Gravity x Height

Given that the efficiency is 85.0%, the height is 160 m, the density of water is 1,000 kg/m³, and the gravity is 9.8 m/s², we need to solve for the flow rate:

2 GW = 0.85 x Flow Rate x 1,000 kg/m³ x 9.8 m/s² x 160 m

Simplifying the equation, we find:

Flow Rate = 2 GW / (0.85 x 1,000 kg/m³ x 9.8 m/s² x 160 m) = 11,920 m³/s

Therefore, approximately 11,920 cubic meters per second of water are needed, assuming 85.0% efficiency.